Set of ordered pairs [closed]












-1












$begingroup$


Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






elementary-set-theory







share|cite|improve this question











$endgroup$



closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, Namaste Dec 25 '18 at 15:06


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, Namaste

If this question can be reworded to fit the rules in the help center, please edit the question.





















    -1












    $begingroup$


    Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






    elementary-set-theory







    share|cite|improve this question











    $endgroup$



    closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, Namaste Dec 25 '18 at 15:06


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, Namaste

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      -1












      -1








      -1





      $begingroup$


      Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






      elementary-set-theory







      share|cite|improve this question











      $endgroup$




      Let $X$ be a set of ordered pairs. How can I show that there exists sets $A$ and $B$ that $Xsubseteq A times B$?






      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 Dec 25 '18 at 11:31









      Asaf Karagila♦

      306k33438769




      306k33438769










      asked Dec 25 '18 at 7:57









      t.ysnt.ysn

      1397




      1397




      closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, Namaste Dec 25 '18 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, Namaste

      If this question can be reworded to fit the rules in the help center, please edit the question.







      closed as off-topic by Saad, Anurag A, Lord_Farin, mrtaurho, Namaste Dec 25 '18 at 15:06


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Anurag A, Lord_Farin, mrtaurho, Namaste

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Just take $A={amidexists b; (a,b)in X}$.



          Any idea of your own now what to take for $B $?






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






            share|cite|improve this answer









            $endgroup$




















              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2












              $begingroup$

              Just take $A={amidexists b; (a,b)in X}$.



              Any idea of your own now what to take for $B $?






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Just take $A={amidexists b; (a,b)in X}$.



                Any idea of your own now what to take for $B $?






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Just take $A={amidexists b; (a,b)in X}$.



                  Any idea of your own now what to take for $B $?






                  share|cite|improve this answer









                  $endgroup$



                  Just take $A={amidexists b; (a,b)in X}$.



                  Any idea of your own now what to take for $B $?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 25 '18 at 8:04









                  drhabdrhab

                  103k545136




                  103k545136























                      1












                      $begingroup$

                      Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.






                          share|cite|improve this answer









                          $endgroup$



                          Let A be the domain of X (first element of every pair in X), and B be the range of X (second element of every pair in X). These do exist, and X is surely the subset of their cartesian product.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 25 '18 at 8:01









                          FrostFrost

                          1776




                          1776















                              -

                              Popular posts from this blog

                              How do I know what Microsoft account the skydrive app is syncing to?

                              Image
                              up vote 4 down vote favorite 1 Some time ago, I set up skydrive for my wife on her mac. But now she wants to access the files online, I cannot remember what Microsoft account I setup and linked her skydive app to and nor can I see anything in the skydrive app settings to tell me either? How can I see the name of the Microsoft account the skydive app is currently linked to? I would like to know this for skydrive app for Mac and PC onedrive share | improve this question asked Feb 25 '13 at 4:52 Gary Barrett 260 2 5 12
                              Read more

                              When does type information flow backwards in C++?

                              Image
                              up vote 62 down vote favorite 19 I just watched Stephan T. Lavavej talk at CppCon 2018 on "Class Template Argument Deduction", where at some point he incidentally says: In C++ type information almost never flows backwards ... I had to say "almost" because there's one or two cases, possibly more but very few . Despite trying to figure out which cases he might be referring to, I couldn't come up with anything. Hence the question: In which cases the C++17 standard mandates that type information propagate backwards? c++ language-lawyer c++17 share | improve this question edited 7 hours ago scohe0
                              Read more

                              Grease: Live!

                              Image
                              A tradução deste artigo está abaixo da qualidade média aceitável. É possível que tenha sido feita por um tradutor automático ou por alguém que não conhece bem o português ou a língua original do texto. Caso queira colaborar com a Wikipédia, tente encontrar a página original e melhore este verbete conforme o guia de tradução. Grease: Live! Grease: Live!  (PRT/BRA)  Estados Unidos 2016 •   cor •   130 min. min   Direção Thomas Kail Alex Rudzinski Produção Marc Platt Adam Siegel Roteiro Allan Carr Robert Cary Warren Casey Jim Jacobs Jonathan Tolins Bronte Woodard Baseado em Grease (musical) Grease (filme) Elenco Julianne Hough Aaron Tveit Vanessa Hudgens Carlos Pena Jr. Keke Palmer Carly Rae Jepsen Kether Donohue Jordan Fisher David Del Rio Andrew Call Gênero Musical Música Tom Kitt Figurino William Ivey Long Companhia(s) produtora(s) Warner Bros. Studios Distribuição Fox Lançamento 31 de Janei
                              Read more