$I_A circ R = R circ I_B = R$ 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







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














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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006902%2fi-a-circ-r-r-circ-i-b-r-where-i-a-a-a-a-in-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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
















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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3006902%2fi-a-circ-r-r-circ-i-b-r-where-i-a-a-a-a-in-a%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more