Prove that if relation $SR$ is symmetric, then $SR = RS$.
$begingroup$
Let $R$ and $S$ be equivalence relations on set $A$.
Prove that if relation $SR$ is symmetric, then $SR = RS$.
$R$ and $S$ are equivalence relations on set $A$, so $Rsubseteq Atimes A$ and $Ssubseteq Atimes A$. In order to be equivalence relations, they need to meet these three conditions:
- be reflexive: $ain A : aRa$ (analogically for $S$)
- be symmetric: $a,bin A : aRb rightarrow bRa$
- be transitive: $a,b,cin A : (aRb wedge bRc)rightarrow aRc$
I do not really know where to start, I was thinking about it the following way:
Composition of relations $S$ and $R$ can be written as $SRsubseteq Atimes A$, as both relations are on set $A$. If relations $S$ and $R$
are equivalence relations, then their composition should also be an
equivalence relation.
but I am not sure whether it is correct and I also get stuck there. It is kind of problematic, as I need a formal proof to it.
elementary-set-theory
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
$endgroup$
add a comment |
$begingroup$
Let $R$ and $S$ be equivalence relations on set $A$.
Prove that if relation $SR$ is symmetric, then $SR = RS$.
$R$ and $S$ are equivalence relations on set $A$, so $Rsubseteq Atimes A$ and $Ssubseteq Atimes A$. In order to be equivalence relations, they need to meet these three conditions:
- be reflexive: $ain A : aRa$ (analogically for $S$)
- be symmetric: $a,bin A : aRb rightarrow bRa$
- be transitive: $a,b,cin A : (aRb wedge bRc)rightarrow aRc$
I do not really know where to start, I was thinking about it the following way:
Composition of relations $S$ and $R$ can be written as $SRsubseteq Atimes A$, as both relations are on set $A$. If relations $S$ and $R$
are equivalence relations, then their composition should also be an
equivalence relation.
but I am not sure whether it is correct and I also get stuck there. It is kind of problematic, as I need a formal proof to it.
elementary-set-theory
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
$endgroup$
$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35
add a comment |
$begingroup$
Let $R$ and $S$ be equivalence relations on set $A$.
Prove that if relation $SR$ is symmetric, then $SR = RS$.
$R$ and $S$ are equivalence relations on set $A$, so $Rsubseteq Atimes A$ and $Ssubseteq Atimes A$. In order to be equivalence relations, they need to meet these three conditions:
- be reflexive: $ain A : aRa$ (analogically for $S$)
- be symmetric: $a,bin A : aRb rightarrow bRa$
- be transitive: $a,b,cin A : (aRb wedge bRc)rightarrow aRc$
I do not really know where to start, I was thinking about it the following way:
Composition of relations $S$ and $R$ can be written as $SRsubseteq Atimes A$, as both relations are on set $A$. If relations $S$ and $R$
are equivalence relations, then their composition should also be an
equivalence relation.
but I am not sure whether it is correct and I also get stuck there. It is kind of problematic, as I need a formal proof to it.
elementary-set-theory
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
$endgroup$
Let $R$ and $S$ be equivalence relations on set $A$.
Prove that if relation $SR$ is symmetric, then $SR = RS$.
$R$ and $S$ are equivalence relations on set $A$, so $Rsubseteq Atimes A$ and $Ssubseteq Atimes A$. In order to be equivalence relations, they need to meet these three conditions:
- be reflexive: $ain A : aRa$ (analogically for $S$)
- be symmetric: $a,bin A : aRb rightarrow bRa$
- be transitive: $a,b,cin A : (aRb wedge bRc)rightarrow aRc$
I do not really know where to start, I was thinking about it the following way:
Composition of relations $S$ and $R$ can be written as $SRsubseteq Atimes A$, as both relations are on set $A$. If relations $S$ and $R$
are equivalence relations, then their composition should also be an
equivalence relation.
but I am not sure whether it is correct and I also get stuck there. It is kind of problematic, as I need a formal proof to it.
elementary-set-theory
elementary-set-theory
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
edited Dec 13 '18 at 6:15
whiskeyo
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
asked Dec 13 '18 at 5:47
whiskeyowhiskeyo
1368
1368
$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35
add a comment |
$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35
$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $SR$ is symmetric then $xSRyiff ySRx$.
We have to show $xSRyiff xRSy$.
So it suffices to show $xRSyiff ySRx$.
So we need to show:
$exists z(xRzSy)iffexists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $Box$
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
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$begingroup$
If $SR$ is symmetric then $xSRyiff ySRx$.
We have to show $xSRyiff xRSy$.
So it suffices to show $xRSyiff ySRx$.
So we need to show:
$exists z(xRzSy)iffexists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $Box$
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
$begingroup$
If $SR$ is symmetric then $xSRyiff ySRx$.
We have to show $xSRyiff xRSy$.
So it suffices to show $xRSyiff ySRx$.
So we need to show:
$exists z(xRzSy)iffexists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $Box$
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
add a comment |
$begingroup$
If $SR$ is symmetric then $xSRyiff ySRx$.
We have to show $xSRyiff xRSy$.
So it suffices to show $xRSyiff ySRx$.
So we need to show:
$exists z(xRzSy)iffexists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $Box$
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
$endgroup$
If $SR$ is symmetric then $xSRyiff ySRx$.
We have to show $xSRyiff xRSy$.
So it suffices to show $xRSyiff ySRx$.
So we need to show:
$exists z(xRzSy)iffexists z(ySzRx)$.
But both of $R$ and $S$ are symmetric, so we can use the same $z$ in both cases. $Box$
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
answered Dec 13 '18 at 6:24
Bjørn Kjos-HanssenBjørn Kjos-Hanssen
2,086918
2,086918
add a comment |
add a comment |
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$begingroup$
My gut instinct would be to use the definition of relation composition, suppose we have $(x,y) in SR$, and then work through the definition and use symmetry to show $(x,y) in RS$ (or, equivalently, $(y,x)$ - not sure which would be more convenient). That would imply the equivalence of the two relations since it would show all elements of $SR$ are shared by $RS$.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 6:08
$begingroup$
How are you defining $SR$? What order of composition?
$endgroup$
– Andrew Li
Dec 13 '18 at 6:21
$begingroup$
If $Psubseteq Atimes B$ and $Qsubseteq Btimes C$, then $PQsubseteq Atimes C$.
$endgroup$
– whiskeyo
Dec 13 '18 at 6:35