Mixing
$I_A circ R = R circ I_B = R$ where $I_A = { (a,a) | a in A }$
Let $R$ be a subset of $Atimes B$. Prove that $I(A) circ R = R circ I(B) = R$, where $I(A) = {(a,a)| a in A}$ .
elementary-set-theory
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
add a comment |
Let $R$ be a subset of $Atimes B$. Prove that $I(A) circ R = R circ I(B) = R$, where $I(A) = {(a,a)| a in A}$ .
elementary-set-theory
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
1
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22
add a comment |
Let $R$ be a subset of $Atimes B$. Prove that $I(A) circ R = R circ I(B) = R$, where $I(A) = {(a,a)| a in A}$ .
elementary-set-theory
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
Let $R$ be a subset of $Atimes B$. Prove that $I(A) circ R = R circ I(B) = R$, where $I(A) = {(a,a)| a in A}$ .
elementary-set-theory
elementary-set-theory
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
edited Nov 20 '18 at 21:22
Eevee Trainer
4,7201634
4,7201634
asked Nov 20 '18 at 21:12
Yahya
106
asked Nov 20 '18 at 21:12
Yahya
106
asked Nov 20 '18 at 21:12
Yahya
106
106
1
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22
add a comment |
1
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22
1
1
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22
add a comment |
1 Answer
1
active
oldest
votes
It's just definitions:
Let $(x,y) in I(A) circ R$. This means by definition that there is some $z in A$ such that $(x,z) in I(A)$ and $(z,y) in R$.
But $(x,z) in I(A)$ means that actually by necessity $z=x$ and so the second fact gives us $(x,y) = (z,y) in R$. So $I(A) circ R subseteq R$
The proof that $R circ I(B) subseteq R$ is similar.
The reverse inclusions are also obvious: if $(x,y) in R$ we can take $z=x$ to note that the pairs $(x,x) in I(A), (x,y)in R$ show that $(x,y) in I(A) circ R$ and similarly for the $I(B)$ case again.
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
add a comment |
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It's just definitions:
Let $(x,y) in I(A) circ R$. This means by definition that there is some $z in A$ such that $(x,z) in I(A)$ and $(z,y) in R$.
But $(x,z) in I(A)$ means that actually by necessity $z=x$ and so the second fact gives us $(x,y) = (z,y) in R$. So $I(A) circ R subseteq R$
The proof that $R circ I(B) subseteq R$ is similar.
The reverse inclusions are also obvious: if $(x,y) in R$ we can take $z=x$ to note that the pairs $(x,x) in I(A), (x,y)in R$ show that $(x,y) in I(A) circ R$ and similarly for the $I(B)$ case again.
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
add a comment |
It's just definitions:
Let $(x,y) in I(A) circ R$. This means by definition that there is some $z in A$ such that $(x,z) in I(A)$ and $(z,y) in R$.
But $(x,z) in I(A)$ means that actually by necessity $z=x$ and so the second fact gives us $(x,y) = (z,y) in R$. So $I(A) circ R subseteq R$
The proof that $R circ I(B) subseteq R$ is similar.
The reverse inclusions are also obvious: if $(x,y) in R$ we can take $z=x$ to note that the pairs $(x,x) in I(A), (x,y)in R$ show that $(x,y) in I(A) circ R$ and similarly for the $I(B)$ case again.
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
add a comment |
It's just definitions:
Let $(x,y) in I(A) circ R$. This means by definition that there is some $z in A$ such that $(x,z) in I(A)$ and $(z,y) in R$.
But $(x,z) in I(A)$ means that actually by necessity $z=x$ and so the second fact gives us $(x,y) = (z,y) in R$. So $I(A) circ R subseteq R$
The proof that $R circ I(B) subseteq R$ is similar.
The reverse inclusions are also obvious: if $(x,y) in R$ we can take $z=x$ to note that the pairs $(x,x) in I(A), (x,y)in R$ show that $(x,y) in I(A) circ R$ and similarly for the $I(B)$ case again.
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
It's just definitions:
Let $(x,y) in I(A) circ R$. This means by definition that there is some $z in A$ such that $(x,z) in I(A)$ and $(z,y) in R$.
But $(x,z) in I(A)$ means that actually by necessity $z=x$ and so the second fact gives us $(x,y) = (z,y) in R$. So $I(A) circ R subseteq R$
The proof that $R circ I(B) subseteq R$ is similar.
The reverse inclusions are also obvious: if $(x,y) in R$ we can take $z=x$ to note that the pairs $(x,x) in I(A), (x,y)in R$ show that $(x,y) in I(A) circ R$ and similarly for the $I(B)$ case again.
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
answered Nov 20 '18 at 21:38
Henno Brandsma
105k347114
105k347114
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
add a comment |
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
In the last section : why we can take x = z?
– Yahya
Nov 20 '18 at 21:49
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 '18 at 21:50
add a comment |
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1
Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
– Eevee Trainer
Nov 20 '18 at 21:13
What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 '18 at 21:18
@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 '18 at 21:19
@Omnomnomnom this is my answer (x,y) ∈ I(A)oR then there is a z : (x,z)∈I(A) and (z,y)∈R by the definition of I(A) : x = z and we have (x,x)∈I(A) and (x,y)∈R I don't have any thing to say
– Yahya
Nov 20 '18 at 21:22