$I_A circ R = R circ I_B = R$ where $I_A = { (a,a) | a in A }$












0














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







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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 at 21:13










  • What are your thoughts on the problem? What have you tried? Where are you getting stuck?
    – Omnomnomnom
    Nov 20 at 21:18










  • @EeveeTrainer I have read this site. But I couldn't prove :(
    – Yahya
    Nov 20 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 at 21:22


















0














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







share|cite|improve this question




















  • 1




    Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
    – Eevee Trainer
    Nov 20 at 21:13










  • What are your thoughts on the problem? What have you tried? Where are you getting stuck?
    – Omnomnomnom
    Nov 20 at 21:18










  • @EeveeTrainer I have read this site. But I couldn't prove :(
    – Yahya
    Nov 20 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 at 21:22
















0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 21:22









Eevee Trainer

4,4131632




4,4131632










asked Nov 20 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 at 21:13










  • What are your thoughts on the problem? What have you tried? Where are you getting stuck?
    – Omnomnomnom
    Nov 20 at 21:18










  • @EeveeTrainer I have read this site. But I couldn't prove :(
    – Yahya
    Nov 20 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 at 21:22
















  • 1




    Proving it should be easy once you understand how to compose relations. en.wikipedia.org/wiki/Composition_of_relations
    – Eevee Trainer
    Nov 20 at 21:13










  • What are your thoughts on the problem? What have you tried? Where are you getting stuck?
    – Omnomnomnom
    Nov 20 at 21:18










  • @EeveeTrainer I have read this site. But I couldn't prove :(
    – Yahya
    Nov 20 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 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 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 at 21:13












What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 at 21:18




What are your thoughts on the problem? What have you tried? Where are you getting stuck?
– Omnomnomnom
Nov 20 at 21:18












@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 at 21:19




@EeveeTrainer I have read this site. But I couldn't prove :(
– Yahya
Nov 20 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 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 at 21:22












1 Answer
1






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • In the last section : why we can take x = z?
    – Yahya
    Nov 20 at 21:49










  • @Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
    – Henno Brandsma
    Nov 20 at 21:50











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1














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.






share|cite|improve this answer





















  • In the last section : why we can take x = z?
    – Yahya
    Nov 20 at 21:49










  • @Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
    – Henno Brandsma
    Nov 20 at 21:50
















1














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.






share|cite|improve this answer





















  • In the last section : why we can take x = z?
    – Yahya
    Nov 20 at 21:49










  • @Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
    – Henno Brandsma
    Nov 20 at 21:50














1












1








1






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 21:38









Henno Brandsma

105k346114




105k346114












  • In the last section : why we can take x = z?
    – Yahya
    Nov 20 at 21:49










  • @Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
    – Henno Brandsma
    Nov 20 at 21:50


















  • In the last section : why we can take x = z?
    – Yahya
    Nov 20 at 21:49










  • @Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
    – Henno Brandsma
    Nov 20 at 21:50
















In the last section : why we can take x = z?
– Yahya
Nov 20 at 21:49




In the last section : why we can take x = z?
– Yahya
Nov 20 at 21:49












@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 at 21:50




@Yahya we take $z=x$ ($z$ is the "connecting element" in the composition)
– Henno Brandsma
Nov 20 at 21:50


















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