Equivalence/(Partial order) relation or any other suitable known relations?Dependence of Axioms of Equivalence Relation?Transitivity of Relations and Eulerian CyclesThe closures of a binary relationNumber of reflexive relations, symmetric relations, reflexive and symmetric relations using digraph approachSet Relations and Equivalence relation.Partial Order Relation and Equivalence RelationsQuotient set and equivalence relationTrue or false? This relation is an equivalence relation: $xRy Leftrightarrow x cdot y$ is evenpairing partial orderings with strict partial orderingsFinding equivalence classes under a relation
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Equivalence/(Partial order) relation or any other suitable known relations?
Dependence of Axioms of Equivalence Relation?Transitivity of Relations and Eulerian CyclesThe closures of a binary relationNumber of reflexive relations, symmetric relations, reflexive and symmetric relations using digraph approachSet Relations and Equivalence relation.Partial Order Relation and Equivalence RelationsQuotient set and equivalence relationTrue or false? This relation is an equivalence relation: $xRy Leftrightarrow x cdot y$ is evenpairing partial orderings with strict partial orderingsFinding equivalence classes under a relation
$begingroup$
Let $G$ be an indirected graph, let $V(G)$ and $E(G)$ be the set of vertices and edges of $G$, respectively. Define a relation $R$ on $G$ as: for all $vin V(G)$ and $ein E(G)$, $vRe$ if and only if $e$ is incident on $v$ and vice versa. Further, considering that $(v, v)=v$ i.e., the self loop $(v, v)$ coincides with the point $v$. Now, i wish to show that $R$ is an equivalence relation or any other suitable known relation on $G$.
My effort: (1) Let $vin V(G)$, then we can say that $v$ is incident on $v$ i.e., $vRv$; but what when $ein E(G)$ is choosen?
(2) Let $vin V(G)$ and $ein E(G)$, then $vReimplies eRv$ or, for any $u, vin V(G)$, $uRvimplies vRu$; but i have no idea of symmetricity when $vin V(G)$ and $ein E(G)$ is choosen or $ein E(G)$ and $fin E(G)$ is choosen? Lastly, transitive law seems to be failed. Please someone suggest me what type of the relation $R$ would be on $G$?
graph-theory relations equivalence-relations
asked Mar 21 at 18:21
getegete
847
$endgroup$
add a comment |
$begingroup$
Let $G$ be an indirected graph, let $V(G)$ and $E(G)$ be the set of vertices and edges of $G$, respectively. Define a relation $R$ on $G$ as: for all $vin V(G)$ and $ein E(G)$, $vRe$ if and only if $e$ is incident on $v$ and vice versa. Further, considering that $(v, v)=v$ i.e., the self loop $(v, v)$ coincides with the point $v$. Now, i wish to show that $R$ is an equivalence relation or any other suitable known relation on $G$.
My effort: (1) Let $vin V(G)$, then we can say that $v$ is incident on $v$ i.e., $vRv$; but what when $ein E(G)$ is choosen?
(2) Let $vin V(G)$ and $ein E(G)$, then $vReimplies eRv$ or, for any $u, vin V(G)$, $uRvimplies vRu$; but i have no idea of symmetricity when $vin V(G)$ and $ein E(G)$ is choosen or $ein E(G)$ and $fin E(G)$ is choosen? Lastly, transitive law seems to be failed. Please someone suggest me what type of the relation $R$ would be on $G$?
graph-theory relations equivalence-relations
asked Mar 21 at 18:21
getegete
847
$endgroup$
$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52
add a comment |
$begingroup$
Let $G$ be an indirected graph, let $V(G)$ and $E(G)$ be the set of vertices and edges of $G$, respectively. Define a relation $R$ on $G$ as: for all $vin V(G)$ and $ein E(G)$, $vRe$ if and only if $e$ is incident on $v$ and vice versa. Further, considering that $(v, v)=v$ i.e., the self loop $(v, v)$ coincides with the point $v$. Now, i wish to show that $R$ is an equivalence relation or any other suitable known relation on $G$.
My effort: (1) Let $vin V(G)$, then we can say that $v$ is incident on $v$ i.e., $vRv$; but what when $ein E(G)$ is choosen?
(2) Let $vin V(G)$ and $ein E(G)$, then $vReimplies eRv$ or, for any $u, vin V(G)$, $uRvimplies vRu$; but i have no idea of symmetricity when $vin V(G)$ and $ein E(G)$ is choosen or $ein E(G)$ and $fin E(G)$ is choosen? Lastly, transitive law seems to be failed. Please someone suggest me what type of the relation $R$ would be on $G$?
graph-theory relations equivalence-relations
asked Mar 21 at 18:21
getegete
847
$endgroup$
Let $G$ be an indirected graph, let $V(G)$ and $E(G)$ be the set of vertices and edges of $G$, respectively. Define a relation $R$ on $G$ as: for all $vin V(G)$ and $ein E(G)$, $vRe$ if and only if $e$ is incident on $v$ and vice versa. Further, considering that $(v, v)=v$ i.e., the self loop $(v, v)$ coincides with the point $v$. Now, i wish to show that $R$ is an equivalence relation or any other suitable known relation on $G$.
My effort: (1) Let $vin V(G)$, then we can say that $v$ is incident on $v$ i.e., $vRv$; but what when $ein E(G)$ is choosen?
(2) Let $vin V(G)$ and $ein E(G)$, then $vReimplies eRv$ or, for any $u, vin V(G)$, $uRvimplies vRu$; but i have no idea of symmetricity when $vin V(G)$ and $ein E(G)$ is choosen or $ein E(G)$ and $fin E(G)$ is choosen? Lastly, transitive law seems to be failed. Please someone suggest me what type of the relation $R$ would be on $G$?
graph-theory relations equivalence-relations
graph-theory relations equivalence-relations
asked Mar 21 at 18:21
getegete
847
asked Mar 21 at 18:21
getegete
847
edited Mar 21 at 18:29
gete
asked Mar 21 at 18:21
getegete
847
asked Mar 21 at 18:21
getegete
847
asked Mar 21 at 18:21
getegete
847
847
$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52
add a comment |
$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52
$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Well, its not an equivalence relation. Its an inhomogeneous relation on $Vtimes E$. For an equivalence relation, you need to start from a homogeneous relation, say a relation on $Vtimes V$.
A prominent equivalence relation on an undirected graph $G$ is defined on $Vtimes V$, with $uRv$ iff there exists a path in $G$ between $u$ and $v$. The equivalence classes are then the connected components of $G$.
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
$endgroup$
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Well, its not an equivalence relation. Its an inhomogeneous relation on $Vtimes E$. For an equivalence relation, you need to start from a homogeneous relation, say a relation on $Vtimes V$.
A prominent equivalence relation on an undirected graph $G$ is defined on $Vtimes V$, with $uRv$ iff there exists a path in $G$ between $u$ and $v$. The equivalence classes are then the connected components of $G$.
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
$endgroup$
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
add a comment |
$begingroup$
Well, its not an equivalence relation. Its an inhomogeneous relation on $Vtimes E$. For an equivalence relation, you need to start from a homogeneous relation, say a relation on $Vtimes V$.
A prominent equivalence relation on an undirected graph $G$ is defined on $Vtimes V$, with $uRv$ iff there exists a path in $G$ between $u$ and $v$. The equivalence classes are then the connected components of $G$.
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
$endgroup$
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
add a comment |
$begingroup$
Well, its not an equivalence relation. Its an inhomogeneous relation on $Vtimes E$. For an equivalence relation, you need to start from a homogeneous relation, say a relation on $Vtimes V$.
A prominent equivalence relation on an undirected graph $G$ is defined on $Vtimes V$, with $uRv$ iff there exists a path in $G$ between $u$ and $v$. The equivalence classes are then the connected components of $G$.
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
$endgroup$
Well, its not an equivalence relation. Its an inhomogeneous relation on $Vtimes E$. For an equivalence relation, you need to start from a homogeneous relation, say a relation on $Vtimes V$.
A prominent equivalence relation on an undirected graph $G$ is defined on $Vtimes V$, with $uRv$ iff there exists a path in $G$ between $u$ and $v$. The equivalence classes are then the connected components of $G$.
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
answered Mar 21 at 18:34
WuestenfuxWuestenfux
5,3631513
5,3631513
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
add a comment |
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
$begingroup$
It looks to me like OP was trying to define the relation on $(Vcup E)times (Vcup E)$.
$endgroup$
– Ethan MacBrough
Mar 21 at 19:12
add a comment |
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$begingroup$
Can you formalize the definition of your relation more clearly? For example it says nothing about whether two edges are ever related to each other. If not the relation is certainly not reflexive.
$endgroup$
– Ethan MacBrough
Mar 21 at 18:34
$begingroup$
@Ethan MacBrough i couldn't define the relation between edges. Could you help me to define one?
$endgroup$
– gete
Mar 21 at 18:43
$begingroup$
That makes no sense. I could definitely the relation on edges arbitrarily. Is your question whether or not there exists a way to define this relation that makes it an equivalence relation?
$endgroup$
– Ethan MacBrough
Mar 21 at 18:52