RoS is a equivalence relation iff RoS = SoR
Let R and S be two equivalence relation on X.
I wanna prove that $Rcirc S$ is an equivalence relation, but I can't prove that it is reflexive and transitive.
For transitivity:
There are arbitrary $x$ and $y$ with $(x,y)in Rcirc S iff$ there is $z$ with $(x,z)in R$ and $(z,y)in S$ ...?
And where I have to use the $Rcirc S=Scirc R$?
elementary-set-theory equivalence-relations
edited Nov 21 '18 at 13:35
Fumera
235
asked Nov 21 '18 at 12:55
Yahya
106
add a comment |
Let R and S be two equivalence relation on X.
I wanna prove that $Rcirc S$ is an equivalence relation, but I can't prove that it is reflexive and transitive.
For transitivity:
There are arbitrary $x$ and $y$ with $(x,y)in Rcirc S iff$ there is $z$ with $(x,z)in R$ and $(z,y)in S$ ...?
And where I have to use the $Rcirc S=Scirc R$?
elementary-set-theory equivalence-relations
edited Nov 21 '18 at 13:35
Fumera
235
asked Nov 21 '18 at 12:55
Yahya
106
What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10
add a comment |
Let R and S be two equivalence relation on X.
I wanna prove that $Rcirc S$ is an equivalence relation, but I can't prove that it is reflexive and transitive.
For transitivity:
There are arbitrary $x$ and $y$ with $(x,y)in Rcirc S iff$ there is $z$ with $(x,z)in R$ and $(z,y)in S$ ...?
And where I have to use the $Rcirc S=Scirc R$?
elementary-set-theory equivalence-relations
edited Nov 21 '18 at 13:35
Fumera
235
asked Nov 21 '18 at 12:55
Yahya
106
Let R and S be two equivalence relation on X.
I wanna prove that $Rcirc S$ is an equivalence relation, but I can't prove that it is reflexive and transitive.
For transitivity:
There are arbitrary $x$ and $y$ with $(x,y)in Rcirc S iff$ there is $z$ with $(x,z)in R$ and $(z,y)in S$ ...?
And where I have to use the $Rcirc S=Scirc R$?
elementary-set-theory equivalence-relations
elementary-set-theory equivalence-relations
edited Nov 21 '18 at 13:35
Fumera
235
asked Nov 21 '18 at 12:55
Yahya
106
edited Nov 21 '18 at 13:35
Fumera
235
asked Nov 21 '18 at 12:55
Yahya
106
edited Nov 21 '18 at 13:35
Fumera
235
edited Nov 21 '18 at 13:35
Fumera
235
edited Nov 21 '18 at 13:35
Fumera
235
235
asked Nov 21 '18 at 12:55
Yahya
106
asked Nov 21 '18 at 12:55
Yahya
106
asked Nov 21 '18 at 12:55
Yahya
106
106
What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10
add a comment |
What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10
What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10
add a comment |
1 Answer
1
active
oldest
votes
I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.
Assumption $Rcirc S=Scirc R$.
Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.
I am confident that you can now prove reflexivity by yourself.
Assumption $Rcirc S$ is an equivalence relation.
This direction is very similar to the proof above, you should give it a try.
answered Nov 21 '18 at 13:21
Fumera
235
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
add a comment |
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I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.
Assumption $Rcirc S=Scirc R$.
Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.
I am confident that you can now prove reflexivity by yourself.
Assumption $Rcirc S$ is an equivalence relation.
This direction is very similar to the proof above, you should give it a try.
answered Nov 21 '18 at 13:21
Fumera
235
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
add a comment |
I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.
Assumption $Rcirc S=Scirc R$.
Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.
I am confident that you can now prove reflexivity by yourself.
Assumption $Rcirc S$ is an equivalence relation.
This direction is very similar to the proof above, you should give it a try.
answered Nov 21 '18 at 13:21
Fumera
235
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
add a comment |
I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.
Assumption $Rcirc S=Scirc R$.
Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.
I am confident that you can now prove reflexivity by yourself.
Assumption $Rcirc S$ is an equivalence relation.
This direction is very similar to the proof above, you should give it a try.
answered Nov 21 '18 at 13:21
Fumera
235
I use the notation $xRy$ instead of $(x,y)in R$. In particular, $xRSy$ means $xRaSy$ for some $a$. It is also convenient to write $xRySz$ instead of ($xRy$ and $yRz$). In general I would advise you do draw such relations as graphs, with vertices representing points, and edges denoting the relationship.
Assumption $Rcirc S=Scirc R$.
Transitivity: Suppose $xRSy$ and $yRSz$, i.e. $xRaSy$ and $yRbSz$ for some $a,b$.
Then $aSyRb$, or short, $aSRb$. Since $SR=RS$, we get $aRSb$, let's say $aRcSb$ for some $c$. But then
$xRaRcSbSz$, and since $R,S$ are equivalence relations, $xRcSz$, so $xRSz$.
I am confident that you can now prove reflexivity by yourself.
Assumption $Rcirc S$ is an equivalence relation.
This direction is very similar to the proof above, you should give it a try.
answered Nov 21 '18 at 13:21
Fumera
235
edited Nov 21 '18 at 13:26
answered Nov 21 '18 at 13:21
Fumera
235
answered Nov 21 '18 at 13:21
Fumera
235
answered Nov 21 '18 at 13:21
Fumera
235
235
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
add a comment |
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
The end of line 2: "But then xRaRcSbSz" I can't understand here. And I have tried to prove reflexivity but I couldn't prove it.
– Yahya
Nov 21 '18 at 13:45
add a comment |
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What does "o" mean here?
– Jean-Claude Arbaut
Nov 21 '18 at 13:06
@Jean-ClaudeArbaut The composition of relatios. look at the definition here: en.wikipedia.org/wiki/Composition_of_relations
– Yahya
Nov 21 '18 at 13:09
I didn't know (or remember) this one. Thanks.
– Jean-Claude Arbaut
Nov 21 '18 at 13:10